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Homework Help: Please help how do i go about solving this?

  1. Feb 6, 2005 #1
    please help how do i go about solving this??

    undefinedundefinedundefined
    A jogger runs in a straight line with an average
    velocity of 4:7 m/s for 5.1 min, and then with
    an average velocity of 4.1 m/s for 3 min.
    What is her total displacement? Answer in
    units of meters
     
  2. jcsd
  3. Feb 6, 2005 #2
    use the formula:
    [tex]d = V_{average}t[/tex]
     
  4. Feb 6, 2005 #3
    and then I subtract the two to find the final displacement
     
  5. Feb 6, 2005 #4

    dextercioby

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    No,you have to add the 2 distances.

    Daniel.
     
  6. Feb 6, 2005 #5
    Well, it all depends what the direction is. If he runs in the same direction both times, you have to add the 2 displacements to find the total displacement.
     
  7. Feb 6, 2005 #6
    thanks dude
     
  8. Feb 6, 2005 #7

    dextercioby

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    Even if the runs in the opposite direction on the second time interval,the TOTAL distance would still be gotten by adding the 2.

    Daniel.
     
  9. Feb 6, 2005 #8
    Aren't the words "distance travelled" and "displacement" different. I though displacement meant distance and direction from the original starting point.
     
  10. Feb 6, 2005 #9
    for the same question, to get the avg. velocity during the time what should i do? i know it is not just to add the 2 velocities given and divide by 2
     
  11. Feb 6, 2005 #10
    Average velocity is total distance divided by total time.
     
  12. Feb 6, 2005 #11

    dextercioby

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    I think that u've given yourself the answer. :wink: Besides,since in the problem we're not given the directions (so that we would think it as a vector projection problem),it's natural to assume the simpler version,which is what the teacher would want,right...? :wink: :tongue2: :biggrin:

    Daniel.
     
  13. Feb 6, 2005 #12
    new question

    A 500-kilogram sports car accelerates uni-
    formly from rest, reaching a speed of 30 me-
    ters per second in 6 seconds.
    During the 6 seconds, the car has traveled
    a distance of how many meters??

    i got 180m is that correct?
    i used : distance = avg. velocity * change in time
    = (30m.s) * 6
    = 180m
     
    Last edited: Feb 6, 2005
  14. Feb 6, 2005 #13
    ha ha, very funny :wink: :biggrin: :tongue:
     
  15. Feb 6, 2005 #14

    dextercioby

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    Check your numbers again...

    Daniel.
     
  16. Feb 6, 2005 #15
    Careful... 30m/s is not the average velocity, but rather the final velocity. Remember, average velocity is total displacement divided by the total time, or in this case (if a is constant),
    [tex]\frac{V_i + V_f}{2} = \frac{0+30m/s}{2}[/tex]
     
  17. Feb 6, 2005 #16

    where did u get the 2 from???
     
  18. Feb 6, 2005 #17
    Isn't the average the sum of the 2 velocities divided by 2? In this case, the displacement would be: [tex]d = V_{ave}t[/tex]
     
  19. Feb 6, 2005 #18

    dextercioby

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    Kay,guys,let's leave side talks and refer to the initial problem ...

    Daniel.
     
  20. Feb 6, 2005 #19
    your saying I take 30m/s and divide it by 2?? that is not right either.
     
  21. Feb 6, 2005 #20
    I's not "side talking". I'm referring to the "new question". Here, average velocity IS involved in finding the distance.

    Anyways, gotta go. I hope I didn't irritate you too much Daniel :biggrin:
     
  22. Feb 6, 2005 #21
    You take the 30 m/s, divide it by 2, and multiply it by the time to get the distance.
     
  23. Feb 6, 2005 #22

    dextercioby

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    Do you know Galileo Galilei's formula for uniformly accelerated motion...??

    Daniel.
     
  24. Feb 6, 2005 #23

    dextercioby

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    Christinono,your method is equivalent to the one i suggested...Let's leave the OP to decide which is better for him to use.
    Daniel.
     
  25. Feb 6, 2005 #24
    galelio

    yes, those are the four different eq'ns
     
  26. Feb 6, 2005 #25

    dextercioby

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    Nope,it's only one.

    [tex]v_{fin}^{2}=v_{init}^{2}+2ad [/tex]

    Daniel.
     
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